Problem: $f(x, y, z) = (xyz, \cos(z), \sin(z))$ What is $\dfrac{\partial f}{\partial z}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(yz, 0, 0)$ (Choice B) B $(xy + yz + zx, -\sin(z), \cos(z))$ (Choice C) C $(xz, 0, 0)$ (Choice D) D $(xy, -\sin(z), \cos(z))$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y, z) = (f_0(x, y, z), f_1(x, y, z), f_2(x, y, z)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x}, \dfrac{\partial f_2}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y}, \dfrac{\partial f_2}{\partial y} \right) \\ \\ &f_z = \left( \dfrac{\partial f_0}{\partial z}, \dfrac{\partial f_1}{\partial z}, \dfrac{\partial f_2}{\partial z} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $z$, we'll treat $x$ and $y$ as if they were constants. Therefore, $f_z = (xy, -\sin(z), \cos(z))$.